Hamming code


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what would the hamming code sequence for the 8 bit sequence 10011101 be? of course this would take up 4 reduntant bits because of the 2^r >= r+m+1 therom... so you have this sequence to figure out the redundant bits

12 ?| 11 | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 ? ? ?positions
1 ? | 0 ?| ?0 | 1 | ? | 1 | 1 | 0 | ? | 1 | ? | ? ? ? ? ?values

as you can see bits bit positions 1,2,4,8 are blank, if you know anything about hamming code could you help me out? i can do this for 7-bit, but 8-bit i forget...

7-bit you find even parity for the following combinations of places

R1 on bits - 1,3,5,7,9,11

R2 on bits - 2,3,6,7,10,11

R4 on bits - 4,5,6,7

R8 on bits - 8,9,10,11

(R just stands for redunant bit, and the number is the position it takes the even parity of the following positions on)

that accounts for 7-bit hamming code with even parity, but not 8-bit hamming codes, which needs to somehow incorporate the 12th bit place holder.

Edited by neufuse
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